Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 10005, M = 29 + 2;
int x[N], y[N], base[M], is_base[N], b[M], sx[M], sy[N];
std::vector<std::pair<int, int> > ans, opt;
inline int read() {
  int now = 0;
  register char c = getchar();
  for (; !isdigit(c); c = getchar())
    ;
  for (; isdigit(c); now = now * 10 + c - 48, c = getchar())
    ;
  return now;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; ++i) x[i] = read();
  for (int i = 1; i <= n; ++i) y[i] = read();
  for (int i = 1; i <= n; ++i) {
    is_base[i] = -1;
    for (int j = 29, s = 0; ~j; --j)
      if (x[i] >> j & 1)
        if (base[j])
          x[i] ^= x[base[j]], s ^= sx[j];
        else {
          is_base[i] = j, base[j] = i, sx[j] = s | (1 << j);
          break;
        }
  }
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    int s = 0;
    for (int j = 29; ~j; --j)
      if (y[i] >> j & 1)
        if (base[j])
          y[i] ^= x[base[j]], s ^= sx[j];
        else
          return puts("-1"), 0;
    if (~is_base[i]) {
      b[cnt] = i, sy[cnt++] = s;
      continue;
    }
    ans.push_back(std::make_pair(i, i));
    for (int j = 29; ~j; --j)
      if (s >> j & 1) ans.push_back(std::make_pair(i, base[j]));
  }
  for (int i = 0; i < cnt; ++i) {
    int s = sy[i];
    sy[i] = 0;
    for (int j = 0; j < cnt; ++j)
      if (s >> is_base[b[j]] & 1) sy[i] |= 1 << j;
  }
  for (int i = 0; i < cnt; ++i) {
    if (!(sy[i] >> i & 1))
      for (int j = i + 1; j < cnt; ++j)
        if (sy[j] >> i & 1) {
          opt.push_back(std::make_pair(b[i], b[j]));
          opt.push_back(std::make_pair(b[j], b[i]));
          opt.push_back(std::make_pair(b[i], b[j]));
          std::swap(sy[i], sy[j]);
          break;
        }
    if (sy[i] >> i & 1)
      for (int j = i + 1; j < cnt; ++j)
        if (sy[j] >> i & 1)
          opt.push_back(std::make_pair(b[j], b[i])), sy[j] ^= sy[i];
  }
  for (int i = 0; i < cnt; ++i) {
    if (!(sy[i] >> i & 1)) ans.push_back(std::make_pair(b[i], b[i]));
    for (int j = i + 1; j < cnt; ++j)
      if (sy[i] >> j & 1) ans.push_back(std::make_pair(b[i], b[j]));
  }
  std::reverse(opt.begin(), opt.end());
  for (auto v : opt) ans.push_back(v);
  printf("%d\n", ans.size());
  for (auto v : ans) printf("%d %d\n", v.first, v.second);
  return 0;
}